Solution to problem 10.11)
Problem 10.20)
**Book has a value for mass of .180kg, and a distance descended of
75.0cm for part b
SWAP .8M FOR .75M, AND YOU GET V_CM=2.711M/S.
DIVIDE BY RADIUS TO FIND ANGULAR VELOCITY OF
33.9RAD/SEC. FINALLY, VELOCITY = OMEGA*R, SO
V=(33.9RAD/SEC*.08m)=2.711M/S
For problem 10.60
Problem 10.86
A small block with mass 0.250 kg is attached to a string
passing through a hole in a frictionless, horizontal surface
(see figure).
The block is originally revolving in a circle with a radius of
0.800 m about the hole with a tangential speed of 4.00 m/s.
The string is then pulled slowly from below, shortening the
radius of the circle in which the block revolves. The breaking
strength of the string is 30.0 N.
What is the radius of the circle when the string breaks ?
Solution
Tension T of the string is radial so that it gives no torque on the block. Angular momemtum is therefore
conserved, i.e., L  mvr  const .
Hence
2
v2 m  L 
L2
T m
 


r
r  mr 
m r3
L  0.250  4.00  0.800  0.800
 0.800
30 
2
0.250 r 3

r
3
 0.800
2
30  0.250
 0.440 m